Suppose a vector $\overrightarrow{A} = A_r\overrightarrow{a_r} + A_\phi\overrightarrow{a_\phi} +A_z\overrightarrow{a_z}$ represents a point $P$ in cylindrical coordinate system and $\overrightarrow{A} = A_x\overrightarrow{a_x} + A_y\overrightarrow{a_y} +A_z\overrightarrow{a_z}$ representing same point $P$ in cartesian system. Please see the picture below for clarity.
So, here comes my question: For locating the point by vector in cartesian form we would move first $A_x$ in $\overrightarrow{a_x}$, $A_y$ in $\overrightarrow{a_y}$ and lastly $A_z$ in $\overrightarrow{a_z}$ and we would reach $P$. But in cylindrical system we can reach $P$ by moving $A_r$ in $\overrightarrow{a_r}$ and we would reach just below $P$ on x-y plane so all we need is to move $A_z$ in $\overrightarrow{a_z}$. So, why do we need the component $A_\phi \overrightarrow{a_\phi}$ for $\overrightarrow{A}$ in cylindrical system? I know I am wrong somewhere so please clear my doubt.
You don't need $a_\theta$ when specifying the position vector since that is already a part of $a_r\hat{r}$. $\vec{r}=r\hat{r}+z\hat{k}$ fully specifies the position vector.
You do need all three basis vectors to specify all the components of a vector located at a position. For example, in circular motion, the velocity of a particle is $\omega r\hat{\theta}$.
A given coordinate implies a basis vector which in turn implies a unit basis vector. As it happens, the definition of cylindrical coordinates implies you don't need a theta component to specify position.
Position vector: $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}=r\cos(\theta)\vec{i}+r\sin(\theta)\vec{j}+z\vec{k}$
Basis vector for $r$ is $\vec{e_r}=\frac{\partial \vec{r}}{\partial r}=\cos(\theta)\vec{i}+\sin(\theta)\vec{j}=\hat{r}$
Basis vector for $\theta$ is $\vec{e_\theta}=\frac{\vec{r}}{\partial \theta} = -r\sin(\theta) \vec{i} + r\cos(\theta)\vec{j}=r\hat{\theta}$ Notice the basis vector has a magnitude of $r$ so $\vec{\theta}=(1/r)\vec{e_\theta}$
For the z coordinate, $\frac{\partial \vec{r}}{\partial z}=\vec{e_z}=1\vec{k}=\hat{k}$.
Re constructing $\vec{r}$ in terms of the unit vectors gives $\vec{r}=r\hat{r}+z\hat{k}$. Notice $\hat{\theta}$ doesn't appear. Further $ (r\hat{r}+z\hat{k})\cdot \hat{\theta}=0$. So there isn't a component of $\vec{r}$ in the direction of $\hat{\theta}$. The angular position and the distance from the z axis are both encoded by $r\hat{r}$.
Note that the unit vectors can be expressed in terms of the unit vectors in cartesian coordinates but the factors vary as functions of $\theta$.
In other words, While no matter where you are in cartesian coordaintes you can use the same constant unit vectors, the unit vectors applicable at a given location vary by that location. That's why for an arbitrary vector, you will often need a theta unit vector.
In circular motion, no matter where on the circle you are, the velocity is $\omega r \hat{\theta}$ because $\hat{\theta}$ implies the proper direction anywhere on the circle despite varying from place to place.